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Diano4ka-milaya [45]
3 years ago
10

A newspaper advisor claims that an advertisement would increase sales by 35%. Sales are currently $500 per month. What will be t

he predicted sale in three years?
Mathematics
1 answer:
dimulka [17.4K]3 years ago
3 0
35% = 0.35
500 x 0.35 = 175
500 + 175 = 675 per month
675 x (12 x 3) = $24,300 sales over three years (I’m not sure what answer is wanted for this question)
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A rectangular solid has sides of 7 cm, 9 cm, and 11 cm. What is its surface area?
otez555 [7]

Answer:

Surface area is, 478 square centimeter

Step-by-step explanation:

Surface area (S)of rectangle is given by:

S = 2(lw+wh+hl)                 .....[1]

where,

l is the length

w is the width

h is the height of the rectangle respectively.

As per the statement:

A rectangular solid has sides of 7 cm, 9 cm, and 11 cm

⇒ l = 7 cm , w = 9 cm and h = 11 cm

Substitute in [1] we have;

S = 2(7 \cdot 9+9 \cdot 11+11 \cdot 7)

⇒S = 2(63+99+77) = 2 \cdot 239 = 478 cm^2

Therefore, the surface area is, 478 square centimeter

7 0
3 years ago
Theoretical probability is approximated by conducting trials and recording the ratio for the number of occurrences of the event
Art [367]
False.
This is the definition of Experimental probability
5 0
3 years ago
20 points and brainly ;) You earned $1600 in interest in your savings account over 10 years at a rate of 8%. How much money did
Mars2501 [29]

A = p(1+r)^t

1600 = p(1+0.08)¹⁰

P = 1600/2.1589

P= 741.12

741.12 dollar money did you originally deposit into your savings account!

7 0
3 years ago
Janie bought 4 apples and 6 bananas. Each apple costs $0.40, and each banana costs $0.60. Enter an expression representing the t
baherus [9]

total cost = c

c =0.40(4) + 0.60(6)

c = 1.60 + 3.60

total cost = 5.20

3 0
3 years ago
The Pew Internet and American Life Project reported Wednesday, April 18th that two- thirds (67%) of young adults with profiles o
dlinn [17]

Answer:

At 5% significance level, larger proportion of military personnel students protect their profiles on social-networking sites than young adults with profiles on social-networking sites.

Step-by-step explanation:

let p be the proportion of military personnel students who restrict access to their profiles. Then null and alternative hypotheses are:

H_{0}: p=0.67 (67%)

H_{a}: p>0.67

We need to calculate z-statistic of sample proportion:

z=\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } } where

  • p(s) is the sample proportion of military students who restrict access to their profiles ( \frac{78}{100} =0.78)
  • p is the proportion assumed under null hypothesis. (0.67)
  • N is the sample size (100)

Then z=\frac{0.78-0.67}{\sqrt{\frac{0.67*0.33}{100} } } ≈ 2.34

The corresponding p-value is 0.0096. Since 0.0096<0.05 (significance level) we can reject the null hypothesis and conclude at 5% significance level that larger proportion of military personnel students protect their profiles on social-networking sites than young adults with profiles on social-networking sites.

3 0
3 years ago
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